how to find the inverse












-1












$begingroup$


gamma



using this equation



$1-sqrt{1-x^2/c^2}$



where c = 1 and x= 0.0 - 1.0 the speed of c



for example



$1-sqrt{1-.886^2/1^2}$ = y = 0.5363147619



gives me the y values on the graph. How do I find the inverse? How do find for x inputting the values of y?



Thank you.






exponential-function inverse







share|cite|improve this question











$endgroup$












  • $begingroup$
    Using the rules of algebra and solving equations ... What have you tried? Remember that $$ frac{1}{(1/b)} = b qquad (bneq 0) $$
    $endgroup$
    – Matti P.
    Jan 14 at 8:13












  • $begingroup$
    You can start by writing $eta = x/c$ and then $$ 1 - frac{1}{(1/sqrt{1-eta^2})} = 1- sqrt{1-eta^2} = y $$ and then move the terms around.
    $endgroup$
    – Matti P.
    Jan 14 at 8:20










  • $begingroup$
    Where is an exponential function ?
    $endgroup$
    – Claude Leibovici
    Jan 14 at 8:53
















-1












$begingroup$


gamma



using this equation



$1-sqrt{1-x^2/c^2}$



where c = 1 and x= 0.0 - 1.0 the speed of c



for example



$1-sqrt{1-.886^2/1^2}$ = y = 0.5363147619



gives me the y values on the graph. How do I find the inverse? How do find for x inputting the values of y?



Thank you.






exponential-function inverse







share|cite|improve this question











$endgroup$












  • $begingroup$
    Using the rules of algebra and solving equations ... What have you tried? Remember that $$ frac{1}{(1/b)} = b qquad (bneq 0) $$
    $endgroup$
    – Matti P.
    Jan 14 at 8:13












  • $begingroup$
    You can start by writing $eta = x/c$ and then $$ 1 - frac{1}{(1/sqrt{1-eta^2})} = 1- sqrt{1-eta^2} = y $$ and then move the terms around.
    $endgroup$
    – Matti P.
    Jan 14 at 8:20










  • $begingroup$
    Where is an exponential function ?
    $endgroup$
    – Claude Leibovici
    Jan 14 at 8:53














-1












-1








-1





$begingroup$


gamma



using this equation



$1-sqrt{1-x^2/c^2}$



where c = 1 and x= 0.0 - 1.0 the speed of c



for example



$1-sqrt{1-.886^2/1^2}$ = y = 0.5363147619



gives me the y values on the graph. How do I find the inverse? How do find for x inputting the values of y?



Thank you.






exponential-function inverse







share|cite|improve this question











$endgroup$




gamma



using this equation



$1-sqrt{1-x^2/c^2}$



where c = 1 and x= 0.0 - 1.0 the speed of c



for example



$1-sqrt{1-.886^2/1^2}$ = y = 0.5363147619



gives me the y values on the graph. How do I find the inverse? How do find for x inputting the values of y?



Thank you.






exponential-function inverse




exponential-function inverse






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 9:51







Agla

















asked Jan 14 at 8:07









AglaAgla

32




32












  • $begingroup$
    Using the rules of algebra and solving equations ... What have you tried? Remember that $$ frac{1}{(1/b)} = b qquad (bneq 0) $$
    $endgroup$
    – Matti P.
    Jan 14 at 8:13












  • $begingroup$
    You can start by writing $eta = x/c$ and then $$ 1 - frac{1}{(1/sqrt{1-eta^2})} = 1- sqrt{1-eta^2} = y $$ and then move the terms around.
    $endgroup$
    – Matti P.
    Jan 14 at 8:20










  • $begingroup$
    Where is an exponential function ?
    $endgroup$
    – Claude Leibovici
    Jan 14 at 8:53


















  • $begingroup$
    Using the rules of algebra and solving equations ... What have you tried? Remember that $$ frac{1}{(1/b)} = b qquad (bneq 0) $$
    $endgroup$
    – Matti P.
    Jan 14 at 8:13












  • $begingroup$
    You can start by writing $eta = x/c$ and then $$ 1 - frac{1}{(1/sqrt{1-eta^2})} = 1- sqrt{1-eta^2} = y $$ and then move the terms around.
    $endgroup$
    – Matti P.
    Jan 14 at 8:20










  • $begingroup$
    Where is an exponential function ?
    $endgroup$
    – Claude Leibovici
    Jan 14 at 8:53
















$begingroup$
Using the rules of algebra and solving equations ... What have you tried? Remember that $$ frac{1}{(1/b)} = b qquad (bneq 0) $$
$endgroup$
– Matti P.
Jan 14 at 8:13






$begingroup$
Using the rules of algebra and solving equations ... What have you tried? Remember that $$ frac{1}{(1/b)} = b qquad (bneq 0) $$
$endgroup$
– Matti P.
Jan 14 at 8:13














$begingroup$
You can start by writing $eta = x/c$ and then $$ 1 - frac{1}{(1/sqrt{1-eta^2})} = 1- sqrt{1-eta^2} = y $$ and then move the terms around.
$endgroup$
– Matti P.
Jan 14 at 8:20




$begingroup$
You can start by writing $eta = x/c$ and then $$ 1 - frac{1}{(1/sqrt{1-eta^2})} = 1- sqrt{1-eta^2} = y $$ and then move the terms around.
$endgroup$
– Matti P.
Jan 14 at 8:20












$begingroup$
Where is an exponential function ?
$endgroup$
– Claude Leibovici
Jan 14 at 8:53




$begingroup$
Where is an exponential function ?
$endgroup$
– Claude Leibovici
Jan 14 at 8:53










2 Answers
2






active

oldest

votes


















1












$begingroup$

Here is the graph for $0le xle 1$



enter image description here



For the inverse, note that $$(x,y)in fiff (y,x)in f^{-1}$$therefore the inverse would be found from $$x=1-sqrt{1-y^2}$$from which we obtain $$f^{-1}(x)=sqrt{2x-x^2}$$






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    If $f(x)=1-frac{1}{frac{1}{sqrt{1-x^2/c^2}}}$, then $f(x)=1-sqrt{1-x^2/c^2}.$



    Hence $sqrt{1-x^2/c^2}=1-f(x)$, thus $1-x^2/c^2=1-2f(x)+f(x)^2.$ Can you proceed ?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      apologies, math was never my strong suit but I will try. Thank you
      $endgroup$
      – Agla
      Jan 14 at 8:34










    • $begingroup$
      sorry I tried. I'm lost
      $endgroup$
      – Agla
      Jan 14 at 9:20











    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Here is the graph for $0le xle 1$



    enter image description here



    For the inverse, note that $$(x,y)in fiff (y,x)in f^{-1}$$therefore the inverse would be found from $$x=1-sqrt{1-y^2}$$from which we obtain $$f^{-1}(x)=sqrt{2x-x^2}$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Here is the graph for $0le xle 1$



      enter image description here



      For the inverse, note that $$(x,y)in fiff (y,x)in f^{-1}$$therefore the inverse would be found from $$x=1-sqrt{1-y^2}$$from which we obtain $$f^{-1}(x)=sqrt{2x-x^2}$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Here is the graph for $0le xle 1$



        enter image description here



        For the inverse, note that $$(x,y)in fiff (y,x)in f^{-1}$$therefore the inverse would be found from $$x=1-sqrt{1-y^2}$$from which we obtain $$f^{-1}(x)=sqrt{2x-x^2}$$






        share|cite|improve this answer









        $endgroup$



        Here is the graph for $0le xle 1$



        enter image description here



        For the inverse, note that $$(x,y)in fiff (y,x)in f^{-1}$$therefore the inverse would be found from $$x=1-sqrt{1-y^2}$$from which we obtain $$f^{-1}(x)=sqrt{2x-x^2}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 14 at 10:20









        Mostafa AyazMostafa Ayaz

        17.2k31039




        17.2k31039























            0












            $begingroup$

            If $f(x)=1-frac{1}{frac{1}{sqrt{1-x^2/c^2}}}$, then $f(x)=1-sqrt{1-x^2/c^2}.$



            Hence $sqrt{1-x^2/c^2}=1-f(x)$, thus $1-x^2/c^2=1-2f(x)+f(x)^2.$ Can you proceed ?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              apologies, math was never my strong suit but I will try. Thank you
              $endgroup$
              – Agla
              Jan 14 at 8:34










            • $begingroup$
              sorry I tried. I'm lost
              $endgroup$
              – Agla
              Jan 14 at 9:20
















            0












            $begingroup$

            If $f(x)=1-frac{1}{frac{1}{sqrt{1-x^2/c^2}}}$, then $f(x)=1-sqrt{1-x^2/c^2}.$



            Hence $sqrt{1-x^2/c^2}=1-f(x)$, thus $1-x^2/c^2=1-2f(x)+f(x)^2.$ Can you proceed ?






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              apologies, math was never my strong suit but I will try. Thank you
              $endgroup$
              – Agla
              Jan 14 at 8:34










            • $begingroup$
              sorry I tried. I'm lost
              $endgroup$
              – Agla
              Jan 14 at 9:20














            0












            0








            0





            $begingroup$

            If $f(x)=1-frac{1}{frac{1}{sqrt{1-x^2/c^2}}}$, then $f(x)=1-sqrt{1-x^2/c^2}.$



            Hence $sqrt{1-x^2/c^2}=1-f(x)$, thus $1-x^2/c^2=1-2f(x)+f(x)^2.$ Can you proceed ?






            share|cite|improve this answer









            $endgroup$



            If $f(x)=1-frac{1}{frac{1}{sqrt{1-x^2/c^2}}}$, then $f(x)=1-sqrt{1-x^2/c^2}.$



            Hence $sqrt{1-x^2/c^2}=1-f(x)$, thus $1-x^2/c^2=1-2f(x)+f(x)^2.$ Can you proceed ?







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 14 at 8:21









            FredFred

            48.6k11849




            48.6k11849












            • $begingroup$
              apologies, math was never my strong suit but I will try. Thank you
              $endgroup$
              – Agla
              Jan 14 at 8:34










            • $begingroup$
              sorry I tried. I'm lost
              $endgroup$
              – Agla
              Jan 14 at 9:20


















            • $begingroup$
              apologies, math was never my strong suit but I will try. Thank you
              $endgroup$
              – Agla
              Jan 14 at 8:34










            • $begingroup$
              sorry I tried. I'm lost
              $endgroup$
              – Agla
              Jan 14 at 9:20
















            $begingroup$
            apologies, math was never my strong suit but I will try. Thank you
            $endgroup$
            – Agla
            Jan 14 at 8:34




            $begingroup$
            apologies, math was never my strong suit but I will try. Thank you
            $endgroup$
            – Agla
            Jan 14 at 8:34












            $begingroup$
            sorry I tried. I'm lost
            $endgroup$
            – Agla
            Jan 14 at 9:20




            $begingroup$
            sorry I tried. I'm lost
            $endgroup$
            – Agla
            Jan 14 at 9:20


















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